$\DeclareMathOperator{\dist}{dist}$ $\newcommand{\bR}{\mathbb{R}}$ $\newcommand{\pa}{\partial}$ Suppose that $N=2$ and $\Omega$ is is the unit disk. Choose $$ u= -1+ar^4+br^5\in C^2(\overline{\Omega}). $$ Then $u=0$ along $\pa \Omega$ implies $a+b=1$. Next $$ \Delta u=\frac{1}{r}\pa_r\big(\; r \pa_r u\;\big)= \frac{1}{r}\pa_r(4ar^4+4br^5)=16ar^2+25br^3. $$ The equality $\Delta u=0$ along $\pa \Omega$ implies $16a+25b=0$. Since $a=1-b$ we deduce $$ 16-16b+25b=0\implies b=-\frac{16}{9},\;\;a=\frac{25}{9}. $$

Note that $$\pa_r u=\frac{1}{9}\big(\; 100 r^3-80 r^4\;\big). $$

Along the boundary we have $\pa_ru=\frac{20}{9}$ which implies that

$$ u(x)\sim \frac{20}{9}\dist\big(x,\Omega\big)\;\;\mbox{near $\pa\Omega$}. $$

$\DeclareMathOperator{\dist}{dist}$ $\newcommand{\bR}{\mathbb{R}}$ $\newcommand{\pa}{\partial}$ Suppose that $N=2$ and $\Omega$ is is the unit disk. Choose $$ u= -1+ar^4+br^5\in C^2(\overline{\Omega}). $$ Then $u=0$ along $\pa \Omega$ implies $a+b=1$. Next $$ \Delta u=\frac{1}{r}\pa_r\big(\; r \pa_r u\;\big)= \frac{1}{r}\pa_r(4ar^4+5br^5)=16ar^2+25br^3. $$ The equality $\Delta u=0$ along $\pa \Omega$ implies $16a+25b=0$. Since $a=1-b$ we deduce $$ 16-16b+25b=0\implies b=-\frac{16}{9},\;\;a=\frac{25}{9}. $$

Note that $$\pa_r u=\frac{1}{9}\big(\; 100 r^3-80 r^4\;\big). $$

Along the boundary we have $\pa_ru=\frac{20}{9}$ which implies that

$$ u(x)\sim \frac{20}{9}\dist\big(x,\Omega\big)\;\;\mbox{near $\pa\Omega$}. $$

$\DeclareMathOperator{\dist}{dist}$ $\newcommand{\bR}{\mathbb{R}}$ $\newcommand{\pa}{\partial}$. Suppose that $N=2$ and $\Omega$ is is the unit disk. Choose $$ u= -1+ar^4+br^5\in C^2(\overline{\Omega}). $$ Then $u=0$ along $\pa \Omega$ implies $a+b=1$. Next $$ \Delta u=\frac{1}{r}\pa_r\big(\; r \pa_r u\;\big)= \frac{1}{r}\pa_r(4ar^4+4br^5)=16ar^2+25br^3. $$ The equality $\Delta u=0$ along $\pa \Omega$ implies $16a+25b=0$. Since $a=1-b$ we deduce $$ 16-16b+25b=0\implies b=-\frac{16}{9},\;\;a=\frac{25}{9}. $$

Note that $$\pa_r u=\frac{1}{9}\big(\; 100 r^3-80 r^4\;\big). $$

Along the boundary we have $\pa_ru=\frac{20}{9}$ which implies that

$$ u(x)\sim \frac{20}{9}\dist\big(x,\Omega\big)\;\;\mbox{near $\pa\Omega$}. $$

$\DeclareMathOperator{\dist}{dist}$ $\newcommand{\bR}{\mathbb{R}}$ $\newcommand{\pa}{\partial}$ Suppose that $N=2$ and $\Omega$ is is the unit disk. Choose $$ u= -1+ar^4+br^5\in C^2(\overline{\Omega}). $$ Then $u=0$ along $\pa \Omega$ implies $a+b=1$. Next $$ \Delta u=\frac{1}{r}\pa_r\big(\; r \pa_r u\;\big)= \frac{1}{r}\pa_r(4ar^4+4br^5)=16ar^2+25br^3. $$ The equality $\Delta u=0$ along $\pa \Omega$ implies $16a+25b=0$. Since $a=1-b$ we deduce $$ 16-16b+25b=0\implies b=-\frac{16}{9},\;\;a=\frac{25}{9}. $$

Note that $$\pa_r u=\frac{1}{9}\big(\; 100 r^3-80 r^4\;\big). $$

Along the boundary we have $\pa_ru=\frac{20}{9}$ which implies that

$$ u(x)\sim \frac{20}{9}\dist\big(x,\Omega\big)\;\;\mbox{near $\pa\Omega$}. $$